Question

3) x² - 2xy-y² = 2, _xy+ y² = 4.​

Answer 1

10²....................

Answer 2

Ответ:

Объяснение:

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Answer 3

To solve the system of equations:

1. x² - 2xy - y² = 2
2. xy + y² = 4

We can use substitution or elimination method. Let's use the substitution method:

From equation (2), we have xy + y² = 4. Rearranging this equation, we get y² + xy - 4 = 0.

Now, let's solve this quadratic equation for y. We can use the quadratic formula:

y = (-x ± √(x² - 4(1)(-4))) / (2(1)) y = (-x ± √(x² + 16)) / 2

Substituting the value of y in equation (1), we get:

x² - 2x((-x ± √(x² + 16)) / 2) - ((-x ± √(x² + 16)) / 2)² = 2

Simplifying this equation will lead to a quadratic equation in terms of x. Let's solve for x:

x² + x√(x² + 16) + (x² + 16)/4 - (x² + 2x√(x² + 16) + (x² + 16)/4) = 2

Simplifying further:

x² + x√(x² + 16) + (x² + 16)/4 - x² - 2x√(x² + 16) - (x² + 16)/4 = 2

Combining like terms:

(x² - x²) + x√(x² + 16) - 2x√(x² + 16) + ((x² + 16)/4 - (x² + 16)/4) = 2

Simplifying:

-x√(x² + 16) = 2

Now, squaring both sides of the equation:

x²(x² + 16) = 4² x⁴ + 16x² = 16 x⁴ + 16x² - 16 = 0

This is a quadratic equation in terms of x². We can solve it by substituting x² as a variable, say u:

u² + 16u - 16 = 0

Now, solve this quadratic equation for u. Once we find the values of u, we can substitute them back to solve for x and y.